APPLIED NUMERICAL METHODS USING MATLAB phần 4 pot

... around the solution (e.g., Fig. 4. 4c or 4. 4d), whereas it con- verges to the solution quickly when f(x) has a steady slope as illustrated in Figs. 4. 4a and 4. 4b. 4. 5 SECANT METHOD The secant method ... “ roots()”. 1.55 1.5 1 .45 1.0000 1.5000 1.3750 1 .42 97 1 .40 77 1 .4 1.35 1 1.2 1 .4 x 2 y = xy = x y = g b ( x ) 1.0000 1.5000 1 .41 67 1 .41 42 1 .41 42 y = g c ( x )...

Ngày tải lên: 09/08/2014, 12:22

... r (3) 2 →     a (4) 11 a (4) 12 a (4) 13 b (4) 1 a (4) 21 a (4) 22 a (4) 23 b (4) 2 a (4) 31 a (4) 32 a (4) 33 b (4) 3     =    2 −1 −10 03/2 −1/21 00 4/ 34/ 3    : r (4) 1 : r (4) 2 : r (4) 3 (2.2.10d) Now, in the stage ... solution as x 3 = b (4) 3 /a (4) 33 = (4/ 3)/ (4/ 3) = 1 x 2 = (b (4) 2 − a (4) 23 x 3 )/a (4) 22 = (1 −(−1/2) × 1)...

Ngày tải lên: 09/08/2014, 12:22

... fminsearch steep fminunc fmincon 1.21 1. 34 1. 34 1. 34 x o 1 0.58 0.62 0.62 0.62 1 f o 0.53 0.17 0.17 0.17 1 −1.21 −1. 34 −1. 34 −1.38 −1.21 −1. 34 −1. 34 −1. 34 1.26 0.00 −0.58 −0.62 −0.62 −0.63 −0.58 ... −1.70 −1.59 1/3 c o −1.76 −1. 34 −1. 34 −1.19 −1.76 −1. 34 −1. 34 −1.33 −1. 84 −0.65 1 −0.00 0.00 0.29 −3.82 −1 .41 −0.00 −0.00 −0.00 −0.00 0.00 −0.00 −0.00 −0.00 −22.1 −16 ....

Ngày tải lên: 09/08/2014, 12:22

... transform, 278, 280, 47 3 Laplace’s Equation, 40 2, 40 4, 42 7, 43 5, 44 2 largest number in MATLAB, 27 leakage, 155, 1 74 least squares (LS), 144 , 165, 169, 171, 351, 3 54 INDEX FOR MATLAB ROUTINES 505 eye() ... sharing via GLOBAL, 44 partial differential equation (PDE), 40 1 partial pivoting, 81,85,105 path, 1 PDE mode, 43 4 PDEtool, 42 9 43 1, 43 5, 45 6 penalty, 346 – 3...

Ngày tải lên: 09/08/2014, 12:22

... procedure for a 4 × 4matrix     a 11 a 12 a 13 a 14 a 12 a 22 a 23 a 24 a 13 a 23 a 33 a 34 a 14 a 24 a 34 a 44     =     u 11 000 u 12 u 22 00 u 13 u 23 u 33 0 u 14 u 24 u 34 u 44         u 11 u 12 u 13 u 14 0 ... u 23 u 22 u 2 13 + u 2 23 + u 2 33 u 13 u 14 + u 23 u 24 + u 33 u 34 u 14 u 11 u 14 u 12 + u 24 u 22 u 14 u 13 + u 24...

Ngày tải lên: 09/08/2014, 12:22

... ||f(x)|| Flops 1 043 1393 x 0 = [1 0.5] x [0.1560 0 .41 11] (P4.7.2) ||f(x)|| 3.97e-15 (3.66e-15) Flops 248 9 3028 x 0 = [1 0.5] x (P4.7.3) ||f(x)|| Flops 147 6 3821 x 0 = [1 0.5] x [0.50 24 0.1506] (P4.7 .4) ||f(x)|| ... 8158 129 64 x 0 = [1 1 1] x [111] (P4.8.2) ||f(x)|| 0 Flops 990 8 54 x 0 = [1 1 1] x (P4.8.3) ||f(x)|| Flops 6611 47 35 x 0 = [1 1 1] x [1.0000 -1.0000 1.0000] (P4.8 .4)...

Ngày tải lên: 09/08/2014, 12:22

... (3 /4) y(0.25) + 1 /4 = 0 .43 75 0.75 y(0.75) = (3 /4) y(0.50) + 1 /4 = 0.5781 1.00 y(1.00) = (1/2)y(0.5) + 1/2 = 3 /4 = 0.75 y(1.00) = (3 /4) y(0.75) + 1 /4 = 0.6836 1.25 y(1.25) = (3 /4) y(1.00) + 1 /4 = ... − 1 and numerical solutions 2 0 24 6810 0 1 3 × 10 4 (a2) Numerical solutions with modifiers 0 246 810 × 10 4 1.5 1 0.5 0 RK4 ABM Hamming (b2) Relative errors with modifie...

Ngày tải lên: 09/08/2014, 12:22

... λy N−1 (P6.11.11b) −−−−−−→ y N−3 − 4y N−2 + 5y N−1 = λy N−1 which can be formulated in a compact form as          5 41 0000 46 41 000 1 46 41 00 0 ·····0 001 46 41 0001 46 4 00001 45                   y 1 y 2 y 3 · y N−3 y N−2 y N−1          = ... 6y i+2 − 4y i+3 + y i +4 = λy i+2 for i = 1:N −5 (P6.11.12) y N 4 − 4y N−3 + 6y N−2 − 4y N−1 + y N = λy N...

Ngày tải lên: 09/08/2014, 12:22

... 0 03/16 1/10 0 007/30 7/ 24 25 /42 /35/6      , p(:, :, 2) =      −1/20 0 −5/8 1/215/16 0 0 05/16 1/20 00−1/2 −5/ 24 0 −5 /40 5/6      , (9 .4. 17) p(:, :, 3) =      4/ 50 0 1/2 6/51/20 ... cos y, u(2,y)=−sin(y) + (3 /4) cos y (P9.2.8) u(x, 0) = 3 /4, u(x,π)=−3 /4 (P9.2.9) Divide the solution region into M x × M y = 20 × 40 sections. (d) 4 ∂ 2 u(x, y) ∂x 2 − 4 ∂ 2...

Ngày tải lên: 09/08/2014, 12:22